It is well known that a ship traveling in a deep water produces wake. Far from the aft of the ship, these wake are called a Kelvin wake. I found the mathematical expression of the shape of the wake, but not its height vs. $r$ where $r$ is the distance from the ship's aft.

  • $\begingroup$ I think there's an answer somewhere in here: "Kelvin Wake Measurements Obtained on Five Surface Ship Models". The PDF is searchable (e.g., for "height"). $\endgroup$
    – Řídící
    Commented Sep 6, 2013 at 14:19
  • $\begingroup$ @Qmechanic and Riccardo: I think the terminology may be slightly off. I think a Kelvin wave is possibly not directly related to a Kelvin wake pattern, which seems what the question is about. $\endgroup$
    – Řídící
    Commented Sep 6, 2013 at 14:26
  • $\begingroup$ @aufkag: Fixed. $\endgroup$
    – Qmechanic
    Commented Sep 6, 2013 at 14:43
  • $\begingroup$ Are you interested in ideal Kelvin wakes (e.g. by a "point ship") or in the physical wakes produced by actual ships? This will heavily influence the types of answer available. $\endgroup$ Commented Sep 6, 2013 at 20:10
  • $\begingroup$ While it stops somewhat short of a full analysis of the height, you might find A Ship's Wake, by Dana Longcope an interesting resource. $\endgroup$ Commented Sep 6, 2013 at 20:17

1 Answer 1


In its section on Integrals with coalescing saddles, the Digital Library of Mathematical Functions provides an exact analysis of the Kelvin wake pattern, starting from the equations derived in A Ship's Wake, by Dana Longcope, and solving them exactly in terms of Airy functions.

Specifically, Longcope represents the water height $z$ at a point $\mathbf x$ using the integral $$ z(\mathbf x,t)=\int A(\mathbf k)e^{i\mathbf k\cdot \mathbf x-i\omega(\mathbf k)t}\text d\mathbf k $$ for some amplitude $A(\mathbf k)$. Using the Doppler-shifted dispersion relation $\omega(\mathbf k)=\sqrt{g |\mathbf k|}-\mathbf u\cdot \mathbf k$ in the ship frame, where $\mathbf u$ is the ship velocity, and asking for stationary waves in the ship frame with $\omega(\mathbf k)=0$, he reduces the wavevector integration to one dimension (i.e. an angle) by enforcing the relation $$\mathbf k=\frac{g}{(\mathbf u\cdot\hat{\mathbf k})^2}\hat{\mathbf k}$$ so that the height is given by $$ z(\mathbf x,t)=\int_0^{2\pi} A(\theta_k) \exp\left[{i\frac{g}{(\mathbf u\cdot\hat{\mathbf k})^2}\hat{\mathbf k}\cdot \mathbf x}\right] \text d\theta_k. \tag 1 $$

This imposes the deep-water dynamics on the wake. The information about the ship is encoded in the amplitudes $A(\theta_k)$. In the DLMF, Berry assumes a point ship radiating backwards: $A$ is constant over $\theta_k\in(-\pi/2,\pi/2)$, and zero outside it. Rewriting the integral in polar coordinates, you obtain $$ z(\phi,\rho)=\int_{{-\pi/2}}^{{\pi/2}}\mathop{\cos}\left(\rho \frac{{\cos}\left(\theta+\phi\right)}{{\cos^{{2}}}\theta}\right)d\theta. \tag 2 $$

Here you have two options. One is to do a stationary-phase analysis of this integral, which will give you simple relations for the shape of the wake, and which if done correctly can also include estimates for the height of the wake. You can also do exact manipulations on the integral and reduce it to a standard form which will involve Airy functions; by then applying the standard asymptotic results for these functions (such as the ones in the DLMF itself) you'll recover the same results as with the first method.

Longcope follows the first method and the DLMF follows the second, and I don't know which one you'll find more appealing. I think they're both too complicated to even state the result here, so I'm going to take the easy way out and claim that (2) is a complete solution to the problem; if you want more details go to either source. Other references you may find interesting are Gravity Waves: The Kelvin Wedge and Related Problems, by Joy Perkinson, or On Kelvin's ship-wave pattern, by F. Ursell (J. Fluid Mech. 8 no. 3 (1960), pp 418-431).

There is one more thing to say, though, and it regards the role of the shape of the ship in determining the height of the wake. All of the results above give relative estimates of the height of the wake: they estimate how the wake dies down with distance and predict the relative heights and depths of the different peaks and troughs, as well as giving the general profile.

However, none of these results can say anything about the absolute height of the wake. One way to think about it is this: think of equation (1) as a kind of "Kelvin transform": the wake creation encodes the shape of the ship and its kinetic energy into the coefficients $A(\mathbf k)$, and from that you extract the wake. You could in principle run this backwards, and use an analysis of the wake to estimate the shape of the ship.

Saying that you approximate your ship as a "point source" means specifying constant coefficients (and predicts the shape of the leading term in the wake) but it doesn't tell you how big those coefficients are in the first place. This is hard to predict and it includes a lot of information:

  • It includes information on the ship's drag. This is because a lot of the ship's kinetic energy is being lost through drag precisely to the creation of the wake. A bigger wake means more energy being lost and thus more drag. Therefore a model that predicts the height of a ship's wake must include hydrodynamic drag as a crucial component.

  • It includes details of the shape of the ship. For example, many ships try to eliminate the wake (and thus drag) by placing a "shadow ship" slightly in front so that its wake will destructively interfere with the ship's. (This is the principle behind bulbous bows which you may have seen and been puzzled by at some point. Thus a model that predicts the height of the wake must incorporate in detail the shape of the ship.

I don't know about you, but to me that sounds hard enough that people could spend their entire university degrees studying this!

Shameless plug: if you've made it this far (or perhaps you haven't), you might take five minutes to see this presentation I gave on Kelvin wakes some time ago.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.